This is a follow-up question to About the limit of the coefficient ratio for a power series over complex numbers.

Cauchy's estimation in complex analysis is a consequence of the Cauchy's integral formula, which can be stated as follow:

Theorem. If $f$ is holomorphic in an open set $\Omega$ that contains a closed disc $D$ centered at $w$ with radius $R$, then for all $n$

$\displaystyle{ \frac{|f^{(n)}(w)|}{n!} \leq \frac{\sup_{z \in C} |f(z)|}{R^n} }$, where $C$ is the boundary circle of $D$.

Note that if $f$ has power series expansion $\sum b_n (z-w)^n$ inside $D$, then $b_n = |f^{(n)}(w)|/n!$.

Consider the following question: If $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc except for a pole at $z_0$ on the unit circle, and $f$ has the power series expansion $\sum a_n z^n$ in the open unit disc. By the answer of Sivaram to the previous post, the coefficients $a_n$ of the expansion of $f$ will not converge to zero. I'm wondering if this can be proved by something that is a (partial) converse to the Cauchy's estimation, and apply it backwards we can ensure that $a_n$ will not go to zero.

I would like to know whether the following version exists:

Assume $f$ is holomorphic in an open set $\Omega$ that contains a closed unit disc except at $z_0$ on the unit circle, and $f$ has power series expansion $\sum a_n z^n$ inside the open unit disc. If for all $n$ large enough and all $R < 1$ we have

$\displaystyle{ |a_n| \leq \frac{\sup_{z \in C} |f(z)|}{R^n} }$,

where $D$ is a closed disc centered at $0$ with radius $R$ and $C$ is the boundary circle of $D$, then $f$ is also holomorphic at $z_0$.

This version may not work at all, but any reasonable modifications which can be used to prove that $a_n$ will be bounded away from zero are fine.

Question. Is there a reasonable version to the converse of the Cauchy's estimation?

  • $\begingroup$ Any function defined by power series is analytic in the interior of the radius of convergence of the series. On the boundary it depends and your $f$ might not even be well defined on the boundary, let alone permit talking about the supremum. Perhaps you would care to rephrase your question? $\endgroup$
    – Aryabhata
    Jan 20, 2011 at 9:11
  • $\begingroup$ @Moron: Sure, I'll try to revise with a better form. $\endgroup$ Jan 20, 2011 at 9:17

2 Answers 2


Unfortunately nothing like that holds. Consider $f(z) = \sqrt{1 - z}$. $f(z)$ is analytic on ${\mathbb C} - \{(x,0): x > 1\}$ and satisfies Cauchy's estimates on circles of radius $r < 1$ since the function is analytic on the unit disc. However, try as you may, you cannot extend $f(z)$ to a neighborhood of $1$. Note that Cauchy's estimates are even satisfied on the circle of radius $1$, by taking limits as $r$ goes to $1$ of the $r < 1$ Cauchy estimates.

  • $\begingroup$ Thank you for the answer, but it seems that $\sqrt{1-z}$ is not holomorphic on any open set that contains the closed unit disc minus a point, or do I miss something? I found that $−\log(1−z)$ serves as a counterexample, even if the constrain on $|a_n|$ is replaced by "$|a_n|$ goes to zero as n approaches infinity". I'll think for a while and try to revise the question again (if it is possible at all). $\endgroup$ Jan 20, 2011 at 17:56
  • $\begingroup$ A branch of $\sqrt{1 - z}$ can be defined on any open set where $\log(1 - z)$ is defined, by defining it to be equal to $e^{{1 \over 2} \log(1 - z)}$. I chose to use the square root function instead of the log because the right-hand side of the Cauchy estimates are finite for the square root function. $\endgroup$
    – Zarrax
    Jan 20, 2011 at 20:21
  • $\begingroup$ You're right about the function $\sqrt{1-z}$, which is definable on any open set as the same as $\log(1-z)$. However I'm still a little confused about whether both these functions can be holomorphic(analytic) on an open set that contains $D = \{z\mid |z|\leq 1\} - \{1\}$, for a fixed branch we selected (say $0 \leq \theta < 2\pi$). (more) $\endgroup$ Jan 21, 2011 at 4:58
  • $\begingroup$ (cont.) For this branch $\sqrt{1-z}$ is not analytic on $\{(x,0)\mid x>1\}$ as you said, and any open set containing $D$ will also contain some non-analytic point besides $1$, and this will violate the first condition which states that $f$ is holomorphic in an open set containing $D$. $\endgroup$ Jan 21, 2011 at 4:59

It is not true. For example, $f(z)=\exp\left(\frac{z+1}{z-1}\right)$ is bounded by $1$ on the unit disk and all the coefficients are also bounded by $1$. But $f$ has an essential singularity at $z=1$.


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